# Arithmetic - Time & Work - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic Arithmetic - Time & Work. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2021 QA Slot 1 | Arithmetic - Time & Work**

Anu, Vinu and Manu can complete a work alone in 15 days, 12 days and 20 days, respectively. Vinu works everyday. Anu works only on alternate days starting from the first day while Manu works only on alternate days starting from the second day. Then, the number of days needed to complete the work is

- A.
8

- B.
6

- C.
5

- D.
7

Answer: Option D

**Explanation** :

Let the total work to be done = LCM (15, 12, 20) = 60 units.

Efficiency of Anu = 60/15 = 4 units/day

Efficiency of Vinu = 60/12 = 5 units/day

Efficiency of Manu = 60/20 = 3 units/day

Let us calculate the work done every 2 days.

Vinu works on both the days, hence work done by Vinu = 2 × 5 = 10 units

Anu works on alternate days, hence work done by Anu = 4 units

Manu works on alternate days, hence work done by Manu = 3 units

∴ Total work done in 2 days = 10 + 4 + 3 = 17 units.

Hence total work done in 6 days = 3 × 17 = 51 units.

On 7th day Vinu and Anu will work, hence work done = 5 + 4 = 9 units.

∴ Work done in total 7 days = 51 + 9 = 60 units [Work gets completed now]

∴ The number of days needed to complete the work is 7 days.

Hence, option (d).

Workspace:

**CAT 2021 QA Slot 1 | Arithmetic - Time & Work**

Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is

Answer: 32

**Explanation** :

Let the total work to be done = LCM (12, 16, 24) = 48 units.

Let the efficiency of Amar, Akbar and Anthony be ‘x’, ‘y’ and ‘z’ units/month

⇒ x + y = 48/12 = 4 …(2)

⇒ y + z = 48/16 = 3 …(1)

⇒ z + x = 48/24 = 2 …(3)

Adding all these equations

⇒ x + y + z = 9/2 = 4.5 …(4)

Solving these four equations we get,

x = 1.5, y = 2.5 and z = 0.5

∴ The person who is neither slowest not fastest is Amar with efficiency of 1.5 units/month.

∴ Time required by Amar to complete the task alone = 48/1.5 = 32 months.

Hence, 32.

Workspace:

**CAT 2021 QA Slot 2 | Arithmetic - Time & Work**

Anil can paint a house in 60 days while Bimal can paint it in 84 days. Anil starts painting and after 10 days, Bimal and Charu join him. Together, they complete the painting in 14 more days. If they are paid a total of ₹ 21000 for the job, then the share of Charu, in INR, proportionate to the work done by him, is

- A.
9100

- B.
9000

- C.
9150

- D.
9200

Answer: Option A

**Explanation** :

Let the total work to be done = LCM (60, 84) = 420 units.

Efficiency of Anil = 420/60 = 7 units/day

Efficiency of Bimal = 420/84 = 5 units/day

Work done by Anil in 10 + 14 days = 24 × 7 = 168 units

Work done by Bimal in 14 days = 14 × 5 = 70 units

∴ Work done by Charu = 420 – 70 - 168 = 182 units

Fraction of work done by Charu = 182/420 = 91/210

∴ Payment received by Charu = 91/210 × 21,000 = Rs. 9,100

Hence, option (a).

Workspace:

**CAT 2021 QA Slot 2 | Arithmetic - Time & Work**

Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is:

- A.
140

- B.
120

- C.
144

- D.
264

Answer: Option C

**Explanation** :

Let the filling and emptying capacity of A and B be ‘a’ and ‘b’ units/hour respectively.

**Case 1**: A is opened at 2 pm and B at 3 pm

Total work done till 10 pm = 8a - 7b

**Case 2**: A is opened at 2 pm and B at 4 pm

Total work done till 6 pm = 4a - 2b

Since work done is same in both cases, we have

8a – 7b = 4a – 2b

⇒ 4a = 5b

Now time taken by A alone to fill the tank = (Total work)/a = (4a – 2b)/a = (5b – 2b)/(5b/4) = 12/5 hours = 144 minutes.

Hence, option (c).

Workspace:

**CAT 2021 QA Slot 3 | Arithmetic - Time & Work**

One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the

- A.
10

- B.
12

- C.
12.5

- D.
11.5

Answer: Option A

**Explanation** :

Let the efficiency of Rahul be ‘r’ units per hour and for Gautam be ‘g’ units per day.

Initially, Rahul works for 8 hours and Gautam works for 6 hours.

∴ Work done by them = 8r + 6g …(1)

If both of them worked starting from 9 AM, they would have completed the work in 7.5 hours.

∴ Work done by them = 7.5r + 7.5g …(2)

⇒ 8r + 6g = 7.5r + 7.5g

⇒ 0.5r = 1.5g

⇒ r = 3g

∴ Work to be done = 8r + 6g = 8r + 2r = 10r

∴ Time taken by Rahul to complete the work alone = 10r/r = 10 hours.

Hence, option (a).

Workspace:

**CAT 2021 QA Slot 3 | Arithmetic - Time & Work**

Anil can paint a house in 12 days while Barun can paint it in 16 days. Anil, Barun, and Chandu undertake to paint the house for ₹ 24000 and the three of them together complete the painting in 6 days. If Chandu is paid in proportion to the work done by him, then the amount in INR received by him is

Answer: 3000

**Explanation** :

Let the area to be painted = LCM(12, 16) = 48 units.

⇒ Efficiency of Anil = 48/12 = 4 units/day

and Efficiency of Barun = 48/16 = 3 units/day

Work done by Anil in 6 days = 6 × 4 = 24 units

Work done by Anil in 6 days = 6 × 3 = 18 units

∴ Remaining work done by Chandu = 48 – 24 – 18 = 6 units.

⇒ Payment received by Chandu = 24000/48 × 6 = Rs. 3,000

**Alternately,**

Fraction of work done by Anil in 6 days = 6/12 = ½

Fraction of work done by Barun in 6 days = 6/16 = 3/8

⇒ Fraction of work done by Chandu in 6 days = 1 - ½ - 3/8 = 1/8

Since, Chandu completes 1/8^{th} of the work, he will receive 1/8^{th} of the payment.

∴ Chandu’s payment = 1/8 × 24,000 = Rs. 3,000

Hence, 3000.

Workspace:

**CAT 2020 QA Slot 2 | Arithmetic - Time & Work**

John takes twice as much time as Jack to finish a job. Jack and Jim together take one-thirds of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than that taken by three of them working together. In how many days will Jim finish the job working alone?

Answer: 4

**Explanation** :

Let John’s efficiency be 1 unit/day.

∴ Jack’s efficiency is 2 units/day.

Jack and Jim’s efficiency is thrice John’s efficienc.

∴ 2 + e_{Jim} = 3 × 1

⇒ e_{Jim} = 1

John takes three days more than that taken by three of them working together

Let the time taken by all three together is d.

∴ Total work to be done = 4 × d = 1 × (d + 3)

⇒ d = 1

∴ Jack finished the work in d + 3 = 4 days.

Since, Jack and Jim have same efficiency, Jim will also finish the work in 4 days.

Hence, 4.

Workspace:

**CAT 2020 QA Slot 3 | Arithmetic - Time & Work**

A contractor agreed to construct a 6 km road in 200 days. He employed 140 persons for the work. After 60 days, he realized that only 1.5 km road has been completed. How many additional people would he need to employ in order to finish the work exactly on time?

Answer: 40

**Explanation** :

In 60 days 1.5 kms out of 6 kms is built.

∴ In 60 days 1.5/6 = 1/4^{th} work is done.

To complete the whole work, it would take 4 × 60 = 240 days, i.e., 180 more days.

Days remaining now is 200 – 60 = 140 days.

∴ To complete the remaining work in 140 days, contractor will have to hire more people. Assuming he hires x more men.

The work which 140 men would take 180 more days, now need to be completed by 140 + x men in 140 days.

∴ 180 × 140 = 140 × (140 + x)

⇒ x = 40

∴ Contractor hires 40 more people.

Hence, 40.

Workspace:

**CAT 2019 QA Slot 1 | Arithmetic - Time & Work**

At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?

- A.
36

- B.
24

- C.
12

- D.
18

Answer: Option D

**Explanation** :

Let total work be the LCM of 12 and 9 = 36 units.

Let the efficiency of A and B be a and b respectively.

Work done per day when A and B are working together = 36/12 = 3 units.

∴ a + b = 3 ...(1)

Work done per day when A is working at half efficieny and B is working at thrice efficiency = 36/9 = 4 units.

∴ (a/2) + 3b = 4 ...(2)

Solving (1) and (2), we get; a = 2.

Time taken by A alone to complete the work = 36/2 = 18 days.

Hence option 4.

Workspace:

**CAT 2019 QA Slot 1 | Arithmetic - Time & Work**

Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

Answer: 13

**Explanation** :

Let the work done by one man and one machine per day be x and y respectively.

Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job.

Since efficiency is inversely proportional to the time taken, so the efficiency of 3 men and 8 machines is twice that of 8 men and 3 machines.

∴ (3x + 8y) = 2(8x + 3y)

∴ 13x = 2y.

So, work done by 13 men in a day = work done by 2 machines in a day.

∴ If two machines can finish the job in 13 days, same work will be done by 13 men in 13 days.

Hence, 13.

Workspace:

**CAT 2019 QA Slot 2 | Arithmetic - Time & Work**

Anil alone can do a job in 20 days while Sunil alone can do it in 40 days. Anil starts the job, and after 3 days, Sunil joins him. Again, after a few more days, Bimal joins them and they together finish the job. If Bimal has done 10% of the job, then in how many days was the job done?

- A.
13

- B.
12

- C.
15

- D.
14

Answer: Option A

**Explanation** :

Anil in one day can do 1/20^{th} of the work

Sunil in one day can do 1/40^{th} of the work

Anil starts the job and Sunil joins him after three days.

So, Anil would have done 3/20^{th} of the work by the time Sunil joins

After Sunil joins, they both would be doing 3/40^{th} of work everyday

Now, it is known that Bimal joins them after some days and finishes 10% of the work (i.e. 1/10^{th }of the work).

Now, Anil alone had done 3/20^{th} of the work in first 3 days and Bimal completes 1/10^{th} of the work

So, in total they would have done 3/20 + 1/10 = 1/4^{th} of the work.

Remaining work = 3/4^{th}, which would be done by Anil and Sunil together.

Anil and Sunil together comple 1/20^{th} + 1/40^{th} = 3/40^{th }work in 1 day.

Hence, time taken by them to comple 3/4^{th} of the work = 10 days.

Combining everything,

Total number of days = 3 + 10 = 13 days.

Hence, option (1).

Workspace:

**CAT 2019 QA Slot 2 | Arithmetic - Time & Work**

John gets Rs 57 per hour of regular work and Rs 114 per hour of overtime work. He works altogether 172 hours and his income from overtime hours is 15% of his income from regular hours. Then, for how many hours did he work overtime?

Answer: 12

**Explanation** :

If John works the same number of regular and over-time hours say 'p'

The income would be 57p and 114p

Let's say that he works 'x' hours regular and 'y' hours overtime...

So, the income would be 57x and 114y

we are told that 114y is 15% of 57x

114y = 0.15 × 57x

y = 0.075x

we also know that x + y = 172

therefore, x + 0.075x = 1.075x = 172

x = 160

y = 172 - 160 = 12

Therefore, the number of hours he worked overtime is 12 hours.

Hence, 12.

Workspace:

**CAT 2018 QA Slot 1 | Arithmetic - Time & Work**

A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?

Answer: 10

**Explanation** :

Let a filling pipe fills the tank at ‘a’ liters per hour and a draining pipe drains at ‘b’ liters per hour.

Therefore,

6(6a – 5b) = 60(5a – 6b)

∴ 6a – 5b = 50a – 60b

∴ 44a = 55b

∴ a = 5b/4

Let the tank gets filled completely in ‘m’ hours when one draining pipe and two filling pipes are on.

∴ m(2a – b) = 6(6a – 5b)

∴ $m\left(2\times \frac{5}{4}b-b\right)$ = $6\left(6\times \frac{5}{4}b-5b\right)$

∴ $m\left(\frac{6b}{4}\right)$ = $6\left(\frac{10b}{4}\right)$

Solving this, we get m = 10

Answer: 10

Workspace:

**CAT 2018 QA Slot 1 | Arithmetic - Time & Work**

Humans and robots can both perform a job but at different efficiencies. Fifteen humans and five robots working together take thirty days to finish the job, whereas five humans and fifteen robots working together take sixty days to finish it. How many days will fifteen humans working together (without any robot) take to finish it?

- A.
36

- B.
32

- C.
45

- D.
40

Answer: Option B

**Explanation** :

Let work-done by a human and a robot in a day are ‘h’ and ‘r’ respectively.

Total work = 30(15h + 5r) = 60(5h + 15r)

∴15h + 5r = 10h + 30r

∴ h = 5r

Let fifteen humans take ‘y’ days to finish the job.

∴ 30(15h + 5r) = y × 15h

∴ 30(15h + h) = y × 15h

∴ y = 32

Hence, option 2.

Workspace:

**CAT 2018 QA Slot 1 | Arithmetic - Time & Work**

When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?

- A.
18

- B.
22

- C.
16

- D.
20

Answer: Option D

**Explanation** :

Ratio of time taken by A and B to complete a certain task = 4 : 5.

∴ Ratio of their efficiencies = 5 : 4.

Now, A completes 50% of the work alone and B completes 5% of the work alone. That means, A and B together complete 45% of the work together in 4 days.

Out of this 45%, work done by A and B individually will be in the ratio of their efficiencies.

∴ Work done by A B in 4 days = 5/9th of 45% = 20%.

⇒ B completes 20% of the work in 4 days.

Thus, B alone can finish the entire job in 20 days.

Hence, option 4.

Workspace:

**CAT 2018 QA Slot 2 | Arithmetic - Time & Work**

A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?

Answer: 48

**Explanation** :

Suppose the inlet pipes of type A fill in water at the rate ‘a’ units per minute and the inlet pipes of type B fill in water at the rate ‘b’ units per minute.

Therefore we have the following

30(10a + 45b) = 60(8a + 18b)

∴ 300a + 1350b = 480a + 1080b

∴ 180a = 270b

∴ a = 1.5b

Total capacity of the tank

= 300a + 1350b = 300(1.5b) + 1350b

= 1800b

If 7 inlet pipes of type A and 27 inlet pipes of type B are opened, the volume of water filled in every minute

= 7a + 27b = 7(1.5b) + 27b = 37.5b

Therefore the number of minutes taken to fill the tank

$=\frac{1800}{37.5}=48$

Hence, 48.

Workspace:

**CAT 2018 QA Slot 2 | Arithmetic - Time & Work**

Ramesh and Ganesh can together complete a work in 16 days. After seven days of working together, Ramesh got sick and his efficiency fell by 30%. As a result, they completed the work in 17 days instead of 16 days. If Ganesh had worked alone after Ramesh got sick, in how many days would he have completed the remaining work?

- A.
12

- B.
11

- C.
13.5

- D.
14.5

Answer: Option C

**Explanation** :

Suppose ‘R’ and ‘G’ are the number of units of work completed by Ramesh and Ganesh everyday. Therefore the total quantum of work to be completed = 16(R + G) = 16R + 16G

For the first 7 days, both work at 100% efficiency and for the remaining 10 days, only Ramesh works at 70% efficiency. Therefore the total quantum of work to be completed = 7(R + G) + 10(0.7R + G) = 14R + 17G

∴ 16R + 16G = 14R + 17G or G = 2R

When Ramesh fell sick, the total quantum of work could have been done in 9 more days with both working at 100% efficiency. Since Ganesh is twice as efficient as Ramesh is, if Ganesh had worked alone, he would have been able to finish the work in

$9\times \frac{3}{2}=13.5$ days

Hence, option 3.

Workspace:

**CAT 2018 QA Slot 2 | Arithmetic - Time & Work**

A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed filling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along?

- A.
4:36 pm

- B.
4:12 pm

- C.
4:48 pm

- D.
4:24 pm

Answer: Option D

**Explanation** :

The time taken by pump A alone to fill the tank completely = t hours.

The time taken by pump B alone to fill the tank completely = ‘t - 2’ hours.

On Wednesday, pump A was used for ‘t-3’ hours and pump B was used for 2 hours to completely fill the tank.

Therefore we have, $\frac{t-3}{t}+\frac{2}{t-2}=1$

$\therefore \; (t\; -\; 3)\; (t\; -\; 2)\; +\; 2t\; =\; t(t\; -\; 2)$

∴ t^{2} - 5t + 6 + 2t = t^{2} - 2t

Solving for t, t = 6.

Therefore pump A takes 6 hours and pump B takes 4 hours to completely fill the tank. Both the tanks are started at 2 pm.

Suppose the capacity of the tank is 12 litres, in one hour A fills 2 litres and B fills 3 litres. Therefore if both pumps A and B are used, they will together fill the tank in

$\frac{12}{2+3}=2.4hours$

= 2 hours and 24 minutes

Therefore on Thursday, the tank will be completely full at 4.24 pm.

Hence, option 4.

Workspace:

**CAT 2017 QA Slot 1 | Arithmetic - Time & Work**

A person can complete a job in 120 days. He works alone on Day 1. On Day 2, he is joined by another person who also can complete the job in exactly 120 days. On Day 3, they are joined by another person of equal efficiency. Like this, everyday a new person with the same efficiency joins the work. How many days are required to complete the job?

Answer: 15

**Explanation** :

Let the number of days required to complete the job be n.

1 person works on day 1, 2 on day 2, 3 on day 3, …. n on day n.

Each person has the same efficiency.

Work = $1\left(\frac{1}{120}\right)+2\left(\frac{1}{120}\right)+3\left(\frac{1}{120}\right)+\cdots +n\left(\frac{1}{120}\right)$

This is also equal to 1.

$\frac{1}{120}+\frac{2}{120}+\frac{3}{120}+..+\frac{n}{120}$ = 1

∑n = 120

n = 15

Hence, 15.

Workspace:

**CAT 2017 QA Slot 2 | Arithmetic - Time & Work**

A tank has an inlet pipe and an outlet pipe. If the outlet pipe is closed then the inlet pipe fills the empty tank in 8 hours. If the outlet pipe is open then the inlet pipe fills the empty tank in 10 hours. If only the outlet pipe is open then in how many hours the full tank becomes half-full?

- A.
20

- B.
30

- C.
40

- D.
45

Answer: Option A

**Explanation** :

In 1 hour, the inlet pipe will fill $\frac{1}{8}$ of the tank.

Let the time taken by the outlet pipe to empty the tank be ‘x’ hours. So in 1 hour,

the outlet pipe will empty $\frac{1}{x}$ of the tank.

Now when the outlet and inlet pipe will empty half the tank in 20 hours.

Hence, option 1.

Workspace:

**CAT 2017 QA Slot 2 | Arithmetic - Time & Work**

Amal can complete a job in 10 days and Bimal can complete it in 8 days. Amal, Bimal and Kamal together complete the job in 4 days and are paid a total amount of Rs 1000 as remuneration. If this amount is shared by them in proportion to their work, then Kamal’s share, in rupees, is

- A.
100

- B.
200

- C.
300

- D.
400

Answer: Option A

**Explanation** :

In one day, Amal, Bimal and Kamal complete $\frac{1}{4}$ of the work.

In one day, Amal and Bimal complete $\frac{1}{10}+\frac{1}{8}=\frac{4+5}{40}=\frac{9}{40}$ of the work.

So amount of work done by Kamal alone in one day is $\frac{1}{4}-\frac{9}{40}$ of the work

Ratio of work done by Amal, Bimal and Kamal in one day $\Rightarrow \frac{1}{10}:\frac{1}{8}:\frac{1}{40}$ or 4 : 5 : 1

Amount received by Kamal $\Rightarrow \frac{1}{(4+5+1)}\times 1000=\frac{1}{10}$ × 1000

= Rs. 10

Hence, option 1.

Workspace:

**CAT 2005 QA | Arithmetic - Time & Work**

A chemical plant has four tanks (A, B, C, and D), each containing 1000 litres of a chemical. The chemical is being pumped from one tank to another as follows:

From A to B @ 20 litres/minute

From C to A @ 90 litres/minute

From A to D @ 10 litres/minute

From C to D @ 50 litres/minute

From B to C @ 100 litres/minute

From D to B @ 110 litres/minute

Which tank gets emptied first and how long does it take (in minutes) to get empty after pumping starts?

- A.
A, 16.66

- B.
C, 20

- C.
D, 20

- D.
D, 25

Answer: Option C

**Explanation** :

The change in the amount of chemical in each tank after every minute is as follows:

A: −20 – 10 + 90 = 60

B: −100 + 110 + 20 = 30

C: −50 − 90 + 100 = −40

D: −110 + 10 + 50 = −50

Since tank D loses the maximum amount of chemical in a minute, it will be emptied first.

Let n minutes be the time taken by tank D to get empty.

∴ 1000 – 50n = 0

∴ n = 20 minutes

Hence, option 3.

Workspace:

**CAT 2004 QA | Arithmetic - Time & Work**

In Nuts And Bolts factory, one machine produces only nuts at the rate of 100 nuts per minute and needs to be cleaned for 5 minutes after production of every 1000 nuts. Another machine produces only bolts at the rate of 75 bolts per minute and needs to be cleaned for 10 minutes after production of every 1500 bolts. If both the machines start production at the same time, what is the minimum duration required for producing 9000 pairs of nuts and bolts?

- A.
130 minutes

- B.
135 minutes

- C.
170 minutes

- D.
180 minutes

Answer: Option C

**Explanation** :

100 nuts are produced per minute.

∴ Time taken to produce 1000 nuts = 10 minutes.

But the machine needs cleaning for 5 minutes after producing 1000 nuts. Thus effective time taken to produce 1000 nuts = 15 minutes.

Effective time taken to produce 9000 nuts = 9 × 15 – time taken for the last cleaning = 135 – 5 = 130 minutes.

75 bolts are produced per minute.

∴ Time taken to produce 1500 bolts = 20 minutes.

But the machine needs cleaning for 10 minutes after producing 1500 bolts. Thus effective time taken to produce 1500 bolts = 30 minutes.

Effective time taken to produce 9000 bolts = 6 × 30 – time taken for the last cleaning = 180 – 10 = 170 minutes.

Thus time taken to produce 9000 pairs of nuts and bolts = 170 minutes.

Hence, option 3.

Workspace:

**CAT 2002 QA | Arithmetic - Time & Work**

It takes 6 technicians a total of 10 hours to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 a.m. and one technician per hour is added beginning at 5 p.m., at what time will the server be complete?

- A.
6:40 p.m.

- B.
7:00 p.m.

- C.
7:20 p.m.

- D.
8:00 p.m.

Answer: Option D

**Explanation** :

6 technicians → 10 hours to 1 job

∴ 1 technician → 60 hours to 1 job

∴ In 6 hrs (11 a.m. - 5 p.m.)6 technicians → ${\frac{3}{5}}^{th}$ of the work.

In the hour after 5p.m.,

7 technicians → ${\frac{7}{60}}^{th}$ of the job.

In the nect hour, 9 technicians will complete ${\frac{9}{60}}^{th}$ of the job.

∴ In 3 hours after 5:00 p.m. → $\frac{24}{60}={\frac{2}{5}}^{th}$ of the job is complete.

∴ At the end of 9^{th} hour i.e. 8:00 p.m. the work is complete.

Hence, option 4.

Workspace:

**CAT 2002 QA | Arithmetic - Time & Work**

Three small pumps and one large pump are filling a tank. Each of the three small pumps works at 2/3rd the rate of the large pump. If all 4 pumps work at the same time, then they should fill the tank in what fraction of time that it would have taken the large pump alone?

- A.
$\frac{4}{7}$

- B.
$\frac{1}{3}$

- C.
$\frac{2}{3}$

- D.
$\frac{3}{4}$

Answer: Option B

**Explanation** :

Work done by the large pump = 3 units

∴ Work done by the 3 small pumps = 2 × 3 = 6 units

∴ Work done by all 4 pumps together = 9 units

$\therefore \frac{\mathrm{Rate}\mathrm{of}\mathrm{work}\mathrm{of}\mathrm{the}\mathrm{large}\mathrm{pump}}{\mathrm{Rate}\mathrm{of}\mathrm{work}\mathrm{of}\mathrm{all}4\mathrm{pumps}}=\frac{3}{9}=\frac{1}{3}$

Rate of work is inversely proportional to time taken for completing the work.

$\therefore \frac{\mathrm{TIme}\mathrm{taken}\mathrm{by}\mathrm{the}\mathrm{large}\mathrm{pump}}{\mathrm{Time}\mathrm{taken}\mathrm{by}\mathrm{all}4\mathrm{pumps}}=\frac{3}{1}$

Hence, option 2.

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